Chapter 2: Pythagoras' Theorem
Pythagoras' Theorem is one of the most fundamental results in mathematics, linking the three sides of any right-angled triangle. In this chapter you will learn to find missing sides in 2D and 3D problems, test whether triangles contain a right angle, and calculate the distance between any two points on a grid.
Exam Board Coverage
Sections 2.1–2.4 and 2.6–2.7 are assessed at both Foundation and Extended tier. Section 2.5 (3D Pythagoras) is predominantly an Extended tier topic for Cambridge 0580 and Edexcel 4MA1, though straightforward 3D questions occasionally appear at Foundation.
2.1 The Theorem and Its Meaning
In any right-angled triangle, the sides have a special relationship. The longest side, which is always opposite the right angle, is called the hypotenuse.
Key Vocabulary
- Hypotenuse — the longest side of a right-angled triangle; it lies opposite the right angle ($90°$)
- Legs (or shorter sides) — the two sides that form the right angle
- Right-angled triangle — a triangle with one interior angle equal to $90°$
Pythagoras' Theorem
In a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$:
$$a^2 + b^2 = c^2$$
The square of the hypotenuse equals the sum of the squares of the two legs.
Right-angled triangle with legs $a$, $b$ and hypotenuse $c$. The right angle is at vertex $A$.
$$a^2 + b^2 = c^2$$
Why Does the Theorem Work?
The most intuitive proof uses areas. Construct a square of side $(a + b)$ and pack it with four copies of the right-angled triangle. The central region that remains is always a square of side $c$, so:
$$\text{Area of large square} = (a+b)^2 = a^2 + 2ab + b^2$$
$$\text{Area of 4 triangles} = 4 \times \tfrac{1}{2}ab = 2ab$$
$$\therefore\; c^2 = (a+b)^2 - 2ab = a^2 + b^2 \quad \checkmark$$
2.2 Finding the Hypotenuse
When you know both shorter sides, use $c^2 = a^2 + b^2$ directly.
Method: Finding the Hypotenuse
- Identify the right angle and label the two legs $a$ and $b$.
- Write $c^2 = a^2 + b^2$.
- Square both known values and add them.
- Take the positive square root: $c = \sqrt{a^2 + b^2}$.
- Round to the required accuracy (usually 3 s.f. or 1 d.p.).
Example 2.1 — Basic hypotenuse
Find the hypotenuse of a right-angled triangle with legs $6\,\text{cm}$ and $8\,\text{cm}$.
Step 1 Label: $a = 6$, $b = 8$, find $c$.
Step 2 $c^2 = 6^2 + 8^2 = 36 + 64 = 100$
Step 3 $c = \sqrt{100} = \mathbf{10\,\text{cm}}$
This is a 6-8-10 Pythagorean triple (a scaled 3-4-5 triangle).
Example 2.2 — Non-integer answer
A right-angled triangle has legs $5\,\text{cm}$ and $9\,\text{cm}$. Find the hypotenuse, giving your answer correct to 3 significant figures.
Step 1 $c^2 = 5^2 + 9^2 = 25 + 81 = 106$
Step 2 $c = \sqrt{106} \approx \mathbf{10.3\,\text{cm}}$ (3 s.f.)
2.3 Finding a Shorter Side
When you know the hypotenuse and one leg, rearrange $c^2 = a^2 + b^2$ to make the unknown leg the subject.
Rearrangement
To find leg $a$: $a^2 = c^2 - b^2$ $\Rightarrow$ $a = \sqrt{c^2 - b^2}$
To find leg $b$: $b^2 = c^2 - a^2$ $\Rightarrow$ $b = \sqrt{c^2 - a^2}$
Memory tip: "Square the hypotenuse, subtract the square of the known side."
Example 2.3 — Finding a shorter side
A right-angled triangle has hypotenuse $13\,\text{cm}$ and one leg $5\,\text{cm}$. Find the other leg.
Step 1 $c = 13$, $b = 5$, find $a$.
Step 2 $a^2 = 13^2 - 5^2 = 169 - 25 = 144$
Step 3 $a = \sqrt{144} = \mathbf{12\,\text{cm}}$
(A 5-12-13 Pythagorean triple.)
Common Mistake
When finding a shorter side, students sometimes write $a^2 = c^2 + b^2$ instead of $c^2 - b^2$. Remember: you are taking the square of the known leg away from the square of the hypotenuse.
2.4 Real-World Applications
Pythagoras' Theorem applies whenever a right angle exists in a physical situation: ladders against walls, diagonals of rectangles, lengths across fields, and navigation problems.
Example 2.4 — Ladder against a wall
A $7\,\text{m}$ ladder leans against a vertical wall. The foot of the ladder is $2.5\,\text{m}$ from the base of the wall. How high up the wall does the ladder reach? Give your answer to 3 s.f.
Step 1 Draw a diagram. The ladder, wall, and ground form a right-angled triangle. The ladder is the hypotenuse ($c = 7$), the ground distance is one leg ($b = 2.5$), and the height $h$ is the other leg.
Step 2 $h^2 = 7^2 - 2.5^2 = 49 - 6.25 = 42.75$
Step 3 $h = \sqrt{42.75} \approx \mathbf{6.54\,\text{m}}$ (3 s.f.)
Example 2.5 — Diagonal of a rectangle
A rectangular playing field is $80\,\text{m}$ long and $45\,\text{m}$ wide. What is the length of a diagonal? Give your answer to the nearest metre.
Step 1 A diagonal splits the rectangle into two right-angled triangles. The two shorter sides are the legs: $a = 80$, $b = 45$.
Step 2 $d^2 = 80^2 + 45^2 = 6400 + 2025 = 8425$
Step 3 $d = \sqrt{8425} \approx 91.8\ldots \approx \mathbf{92\,\text{m}}$
2.5 3D Pythagoras Extended
In three-dimensional problems, Pythagoras' Theorem is applied in two stages. The key idea is to identify a right-angled triangle within the 3D shape and use it to find the length you need.
Two-Step Method for 3D Pythagoras
- First triangle: Identify a 2D right-angled triangle in the base (or another face). Calculate the required diagonal or length using $c^2 = a^2 + b^2$.
- Second triangle: Use this length as one side of a second right-angled triangle that includes the 3D length you want.
Example 2.6 — Space diagonal of a cuboid
A cuboid has length $12\,\text{cm}$, width $5\,\text{cm}$, and height $8\,\text{cm}$. Find the length of the space diagonal (corner to opposite corner). Give your answer to 3 s.f.
Step 1 Find the base diagonal $d$: $$d^2 = 12^2 + 5^2 = 144 + 25 = 169 \implies d = 13\,\text{cm}$$
Step 2 The space diagonal $D$ forms a right-angled triangle with $d = 13$ and height $h = 8$: $$D^2 = 13^2 + 8^2 = 169 + 64 = 233$$
Step 3 $D = \sqrt{233} \approx \mathbf{15.3\,\text{cm}}$ (3 s.f.)
Other 3D Shapes
The same two-step approach works for:
- Square-based pyramid: Find the diagonal of the base first, then use half of it to find the slant edge or the perpendicular height.
- Cone: The slant height $l$, perpendicular height $h$, and radius $r$ satisfy $l^2 = h^2 + r^2$. This is a single-step application.
2.6 Converse of Pythagoras' Theorem
The converse is a powerful tool: it lets you test whether a triangle is right-angled, without measuring angles directly.
Converse of Pythagoras' Theorem
Let a triangle have sides of length $p$, $q$, and $r$, where $r$ is the longest side.
- If $p^2 + q^2 = r^2$, then the triangle contains a right angle (at the vertex opposite side $r$).
- If $p^2 + q^2 > r^2$, then all angles are acute.
- If $p^2 + q^2 < r^2$, then one angle is obtuse.
Example 2.7 — Converse test
Determine whether a triangle with sides $7\,\text{cm}$, $24\,\text{cm}$, $25\,\text{cm}$ is right-angled.
Step 1 Identify the longest side: $r = 25$. Remaining sides: $p = 7$, $q = 24$.
Step 2 Test: $p^2 + q^2 = 49 + 576 = 625$
Step 3 Compare: $r^2 = 25^2 = 625$
Conclusion Since $p^2 + q^2 = r^2$, the triangle is right-angled (7-24-25 is a Pythagorean triple).
Exam Tip
Always square the longest side and compare it with the sum of the squares of the other two. If you square the wrong side you will get the wrong conclusion.
Common Pythagorean Triples
These integer triples satisfy $a^2 + b^2 = c^2$ exactly and appear frequently in exam questions:
| Triple $(a, b, c)$ | Common multiples |
|---|---|
| 3, 4, 5 | 6-8-10, 9-12-15, 15-20-25, … |
| 5, 12, 13 | 10-24-26, … |
| 8, 15, 17 | 16-30-34, … |
| 7, 24, 25 | 14-48-50, … |
2.7 Distance Between Two Points
When two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ are plotted on a coordinate grid, you can find the distance between them by treating the horizontal and vertical separations as the legs of a right-angled triangle.
Distance Formula
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
This is simply Pythagoras' Theorem with legs $|x_2 - x_1|$ (horizontal) and $|y_2 - y_1|$ (vertical).
Example 2.8 — Distance between two points
Find the distance between $A(1, 2)$ and $B(7, 10)$.
Step 1 Horizontal gap: $x_2 - x_1 = 7 - 1 = 6$
Step 2 Vertical gap: $y_2 - y_1 = 10 - 2 = 8$
Step 3 $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$
Example 2.9 — Distance with negative coordinates
Calculate the distance between $C(-3, 4)$ and $D(5, -2)$. Give your answer to 3 s.f.
Step 1 $x_2 - x_1 = 5 - (-3) = 8$
Step 2 $y_2 - y_1 = -2 - 4 = -6$
Step 3 $d = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10.0}$
Note: $(-6)^2 = 36$ — squaring removes the negative sign automatically.
Negative Differences Are Fine
Since we square the differences, a negative value for $(x_2 - x_1)$ or $(y_2 - y_1)$ causes no problem. You can also compute $(x_1 - x_2)^2$ — the answer is identical.
Exercises
Attempt each question before revealing the solution. Where a decimal answer is required, give it to 3 significant figures unless stated otherwise.
Exercise 2.1
Find the hypotenuse of a right-angled triangle with legs $9\,\text{cm}$ and $40\,\text{cm}$.
Show Solution
$c^2 = 9^2 + 40^2 = 81 + 1600 = 1681$
$c = \sqrt{1681} = \mathbf{41\,\text{cm}}$
(9-40-41 is a Pythagorean triple.)
Exercise 2.2
A right-angled triangle has hypotenuse $17\,\text{m}$ and one leg $8\,\text{m}$. Find the length of the other leg.
Show Solution
$a^2 = 17^2 - 8^2 = 289 - 64 = 225$
$a = \sqrt{225} = \mathbf{15\,\text{m}}$
Exercise 2.3
A ramp rises $1.8\,\text{m}$ vertically over a horizontal distance of $4.2\,\text{m}$. Find the length of the ramp (the slant surface) to 3 s.f.
Show Solution
$c^2 = 1.8^2 + 4.2^2 = 3.24 + 17.64 = 20.88$
$c = \sqrt{20.88} \approx \mathbf{4.57\,\text{m}}$
Exercise 2.4
A rectangle has a diagonal of length $26\,\text{cm}$ and one side of length $10\,\text{cm}$. Find the length of the other side.
Show Solution
$b^2 = 26^2 - 10^2 = 676 - 100 = 576$
$b = \sqrt{576} = \mathbf{24\,\text{cm}}$
Exercise 2.5
A ship sails $15\,\text{km}$ due East, then $20\,\text{km}$ due North. Find the straight-line distance from its starting point.
Show Solution
East and North are perpendicular, so Pythagoras applies.
$d^2 = 15^2 + 20^2 = 225 + 400 = 625$
$d = \sqrt{625} = \mathbf{25\,\text{km}}$
Exercise 2.6
Determine whether a triangle with sides $11\,\text{cm}$, $60\,\text{cm}$, $61\,\text{cm}$ is right-angled. Show all working.
Show Solution
Longest side: $r = 61$. Check: $p^2 + q^2 = 11^2 + 60^2 = 121 + 3600 = 3721$
$r^2 = 61^2 = 3721$ ✓
Since $p^2 + q^2 = r^2$, the triangle is right-angled.
Exercise 2.7
Find the distance between $P(2, -1)$ and $Q(8, 7)$.
Show Solution
Horizontal: $8 - 2 = 6$; Vertical: $7 - (-1) = 8$
$d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$
Exercise 2.8
Find the distance between $A(-5, 3)$ and $B(2, -4)$, giving your answer to 3 s.f.
Show Solution
Horizontal: $2 - (-5) = 7$; Vertical: $-4 - 3 = -7$
$d = \sqrt{7^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} \approx \mathbf{9.90}$
Exercise 2.9 Extended
A cuboid has dimensions $6\,\text{cm} \times 8\,\text{cm} \times 10\,\text{cm}$. Calculate the length of the space diagonal to 3 s.f.
Show Solution
Step 1 — base diagonal: $d^2 = 6^2 + 8^2 = 36 + 64 = 100 \Rightarrow d = 10$
Step 2 — space diagonal: $D^2 = 10^2 + 10^2 = 100 + 100 = 200$
$D = \sqrt{200} \approx \mathbf{14.1\,\text{cm}}$
Exercise 2.10 Extended
A cone has a slant height of $15\,\text{cm}$ and a base radius of $9\,\text{cm}$. Find the perpendicular height of the cone to 3 s.f.
Show Solution
The slant height $l$, perpendicular height $h$, and radius $r$ form a right-angled triangle:
$h^2 = l^2 - r^2 = 15^2 - 9^2 = 225 - 81 = 144$
$h = \sqrt{144} = \mathbf{12\,\text{cm}}$
Exam Tips
Tip 1 — Identify the hypotenuse first
Before writing any formula, mark the right angle and then label the side opposite it as the hypotenuse $c$. This prevents substituting into the wrong position in the equation.
Tip 2 — Always square before adding or subtracting
Write $c^2 = a^2 + b^2$ as a single line, showing each individual square:
Good: $c^2 = 5^2 + 12^2 = 25 + 144 = 169$
Bad: $c^2 = 5 + 12 = 17$
Tip 3 — Show the square root step separately
After computing $c^2 = 169$, write $c = \sqrt{169} = 13$ on the next line. Examiners award a method mark for this step even if you make an arithmetic error earlier.
Tip 4 — 3D problems: draw two separate triangles
Sketch the first right-angled triangle (inside the base), find that diagonal, then sketch a second triangle showing the vertical dimension. Treating them as two separate Pythagoras calculations prevents errors.
Tip 5 — Check your answer makes sense
The hypotenuse must be the longest side. If your answer for the hypotenuse is shorter than one of the legs, you have made an error — go back and check.